3.150 \(\int \frac {(d+e x^2)^4}{\sqrt {a+c x^4}} \, dx\)

Optimal. Leaf size=388 \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (-252 a^{3/2} \sqrt {c} d e^3+25 a^2 e^4+420 \sqrt {a} c^{3/2} d^3 e-210 a c d^2 e^2+105 c^2 d^4\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{210 \sqrt [4]{a} c^{9/4} \sqrt {a+c x^4}}+\frac {4 d e x \sqrt {a+c x^4} \left (5 c d^2-3 a e^2\right )}{5 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {4 \sqrt [4]{a} d e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (5 c d^2-3 a e^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {a+c x^4}}+\frac {e^2 x \sqrt {a+c x^4} \left (42 c d^2-5 a e^2\right )}{21 c^2}+\frac {4 d e^3 x^3 \sqrt {a+c x^4}}{5 c}+\frac {e^4 x^5 \sqrt {a+c x^4}}{7 c} \]

[Out]

1/21*e^2*(-5*a*e^2+42*c*d^2)*x*(c*x^4+a)^(1/2)/c^2+4/5*d*e^3*x^3*(c*x^4+a)^(1/2)/c+1/7*e^4*x^5*(c*x^4+a)^(1/2)
/c+4/5*d*e*(-3*a*e^2+5*c*d^2)*x*(c*x^4+a)^(1/2)/c^(3/2)/(a^(1/2)+x^2*c^(1/2))-4/5*a^(1/4)*d*e*(-3*a*e^2+5*c*d^
2)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*
x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^4+a)^(1/
2)+1/210*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^
(1/4)*x/a^(1/4))),1/2*2^(1/2))*(105*c^2*d^4-210*a*c*d^2*e^2+25*a^2*e^4+420*c^(3/2)*d^3*e*a^(1/2)-252*a^(3/2)*d
*e^3*c^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/c^(9/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.42, antiderivative size = 386, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1207, 1888, 1198, 220, 1196} \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (5 \left (5 a^2 e^4-42 a c d^2 e^2+21 c^2 d^4\right )+84 \sqrt {a} \sqrt {c} d e \left (5 c d^2-3 a e^2\right )\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{210 \sqrt [4]{a} c^{9/4} \sqrt {a+c x^4}}+\frac {e^2 x \sqrt {a+c x^4} \left (42 c d^2-5 a e^2\right )}{21 c^2}+\frac {4 d e x \sqrt {a+c x^4} \left (5 c d^2-3 a e^2\right )}{5 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {4 \sqrt [4]{a} d e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (5 c d^2-3 a e^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {a+c x^4}}+\frac {4 d e^3 x^3 \sqrt {a+c x^4}}{5 c}+\frac {e^4 x^5 \sqrt {a+c x^4}}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)^4/Sqrt[a + c*x^4],x]

[Out]

(e^2*(42*c*d^2 - 5*a*e^2)*x*Sqrt[a + c*x^4])/(21*c^2) + (4*d*e^3*x^3*Sqrt[a + c*x^4])/(5*c) + (e^4*x^5*Sqrt[a
+ c*x^4])/(7*c) + (4*d*e*(5*c*d^2 - 3*a*e^2)*x*Sqrt[a + c*x^4])/(5*c^(3/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (4*a^(1/
4)*d*e*(5*c*d^2 - 3*a*e^2)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*Arc
Tan[(c^(1/4)*x)/a^(1/4)], 1/2])/(5*c^(7/4)*Sqrt[a + c*x^4]) + ((84*Sqrt[a]*Sqrt[c]*d*e*(5*c*d^2 - 3*a*e^2) + 5
*(21*c^2*d^4 - 42*a*c*d^2*e^2 + 5*a^2*e^4))*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2
]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(210*a^(1/4)*c^(9/4)*Sqrt[a + c*x^4])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rule 1207

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e^q*x^(2*q - 3)*(a + c*x^4)^(p +
1))/(c*(4*p + 2*q + 1)), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + c*x^4)^p*ExpandToSum[c*(4*p + 2*q + 1)*(d
+ e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, c, d, e, p},
x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[q, 1]

Rule 1888

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, D
ist[1/(b*(q + n*p + 1)), Int[ExpandToSum[b*(q + n*p + 1)*(Pq - Pqq*x^q) - a*Pqq*(q - n + 1)*x^(q - n), x]*(a +
b*x^n)^p, x], x] + Simp[(Pqq*x^(q - n + 1)*(a + b*x^n)^(p + 1))/(b*(q + n*p + 1)), x]] /; NeQ[q + n*p + 1, 0]
&& q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IG
tQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^4}{\sqrt {a+c x^4}} \, dx &=\frac {e^4 x^5 \sqrt {a+c x^4}}{7 c}+\frac {\int \frac {7 c d^4+28 c d^3 e x^2+e^2 \left (42 c d^2-5 a e^2\right ) x^4+28 c d e^3 x^6}{\sqrt {a+c x^4}} \, dx}{7 c}\\ &=\frac {4 d e^3 x^3 \sqrt {a+c x^4}}{5 c}+\frac {e^4 x^5 \sqrt {a+c x^4}}{7 c}+\frac {\int \frac {35 c^2 d^4+28 c d e \left (5 c d^2-3 a e^2\right ) x^2+5 c e^2 \left (42 c d^2-5 a e^2\right ) x^4}{\sqrt {a+c x^4}} \, dx}{35 c^2}\\ &=\frac {e^2 \left (42 c d^2-5 a e^2\right ) x \sqrt {a+c x^4}}{21 c^2}+\frac {4 d e^3 x^3 \sqrt {a+c x^4}}{5 c}+\frac {e^4 x^5 \sqrt {a+c x^4}}{7 c}+\frac {\int \frac {5 c \left (21 c^2 d^4-42 a c d^2 e^2+5 a^2 e^4\right )+84 c^2 d e \left (5 c d^2-3 a e^2\right ) x^2}{\sqrt {a+c x^4}} \, dx}{105 c^3}\\ &=\frac {e^2 \left (42 c d^2-5 a e^2\right ) x \sqrt {a+c x^4}}{21 c^2}+\frac {4 d e^3 x^3 \sqrt {a+c x^4}}{5 c}+\frac {e^4 x^5 \sqrt {a+c x^4}}{7 c}-\frac {\left (4 \sqrt {a} d e \left (5 c d^2-3 a e^2\right )\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{5 c^{3/2}}+\frac {\left (105 c^2 d^4+420 \sqrt {a} c^{3/2} d^3 e-210 a c d^2 e^2-252 a^{3/2} \sqrt {c} d e^3+25 a^2 e^4\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{105 c^2}\\ &=\frac {e^2 \left (42 c d^2-5 a e^2\right ) x \sqrt {a+c x^4}}{21 c^2}+\frac {4 d e^3 x^3 \sqrt {a+c x^4}}{5 c}+\frac {e^4 x^5 \sqrt {a+c x^4}}{7 c}+\frac {4 d e \left (5 c d^2-3 a e^2\right ) x \sqrt {a+c x^4}}{5 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {4 \sqrt [4]{a} d e \left (5 c d^2-3 a e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {a+c x^4}}+\frac {\left (105 c^2 d^4+420 \sqrt {a} c^{3/2} d^3 e-210 a c d^2 e^2-252 a^{3/2} \sqrt {c} d e^3+25 a^2 e^4\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{210 \sqrt [4]{a} c^{9/4} \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 203, normalized size = 0.52 \[ \frac {e x \left (-25 a^2 e^3+28 c d x^2 \sqrt {\frac {c x^4}{a}+1} \left (5 c d^2-3 a e^2\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^4}{a}\right )+2 a c e \left (105 d^2+42 d e x^2-5 e^2 x^4\right )+3 c^2 e x^4 \left (70 d^2+28 d e x^2+5 e^2 x^4\right )\right )+5 x \sqrt {\frac {c x^4}{a}+1} \left (5 a^2 e^4-42 a c d^2 e^2+21 c^2 d^4\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right )}{105 c^2 \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)^4/Sqrt[a + c*x^4],x]

[Out]

(5*(21*c^2*d^4 - 42*a*c*d^2*e^2 + 5*a^2*e^4)*x*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/
a)] + e*x*(-25*a^2*e^3 + 2*a*c*e*(105*d^2 + 42*d*e*x^2 - 5*e^2*x^4) + 3*c^2*e*x^4*(70*d^2 + 28*d*e*x^2 + 5*e^2
*x^4) + 28*c*d*(5*c*d^2 - 3*a*e^2)*x^2*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^4)/a)]))/(1
05*c^2*Sqrt[a + c*x^4])

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{4} x^{8} + 4 \, d e^{3} x^{6} + 6 \, d^{2} e^{2} x^{4} + 4 \, d^{3} e x^{2} + d^{4}}{\sqrt {c x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^4/(c*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

integral((e^4*x^8 + 4*d*e^3*x^6 + 6*d^2*e^2*x^4 + 4*d^3*e*x^2 + d^4)/sqrt(c*x^4 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{4}}{\sqrt {c x^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^4/(c*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^4/sqrt(c*x^4 + a), x)

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maple [C]  time = 0.02, size = 506, normalized size = 1.30 \[ \frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, d^{4} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {4 i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {a}\, d^{3} e}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+6 \left (-\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, c}+\frac {\sqrt {c \,x^{4}+a}\, x}{3 c}\right ) d^{2} e^{2}+4 \left (\frac {\sqrt {c \,x^{4}+a}\, x^{3}}{5 c}-\frac {3 i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) a^{\frac {3}{2}}}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, c^{\frac {3}{2}}}\right ) d \,e^{3}+\left (\frac {\sqrt {c \,x^{4}+a}\, x^{5}}{7 c}+\frac {5 \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a^{2} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{21 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, c^{2}}-\frac {5 \sqrt {c \,x^{4}+a}\, a x}{21 c^{2}}\right ) e^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^4/(c*x^4+a)^(1/2),x)

[Out]

e^4*(1/7/c*x^5*(c*x^4+a)^(1/2)-5/21*a/c^2*x*(c*x^4+a)^(1/2)+5/21*a^2/c^2/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)
*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)
)+4*d*e^3*(1/5/c*x^3*(c*x^4+a)^(1/2)-3/5*I*a^(3/2)/c^(3/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1
)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*(EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)-EllipticE((I
/a^(1/2)*c^(1/2))^(1/2)*x,I)))+6*d^2*e^2*(1/3*(c*x^4+a)^(1/2)/c*x-1/3*a/c/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2
)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I
))+4*I*d^3*e*a^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2
)/(c*x^4+a)^(1/2)/c^(1/2)*(EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)-EllipticE((I/a^(1/2)*c^(1/2))^(1/2)*x,I))+
d^4/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)
*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{4}}{\sqrt {c x^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^4/(c*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^4/sqrt(c*x^4 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x^2+d\right )}^4}{\sqrt {c\,x^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^2)^4/(a + c*x^4)^(1/2),x)

[Out]

int((d + e*x^2)^4/(a + c*x^4)^(1/2), x)

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sympy [C]  time = 6.17, size = 214, normalized size = 0.55 \[ \frac {d^{4} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {d^{3} e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{\sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {3 d^{2} e^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {d e^{3} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{\sqrt {a} \Gamma \left (\frac {11}{4}\right )} + \frac {e^{4} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**4/(c*x**4+a)**(1/2),x)

[Out]

d**4*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4)) + d**3*e*x**3*gam
ma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(sqrt(a)*gamma(7/4)) + 3*d**2*e**2*x**5*gamma(5/4)
*hyper((1/2, 5/4), (9/4,), c*x**4*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(9/4)) + d*e**3*x**7*gamma(7/4)*hyper((1/
2, 7/4), (11/4,), c*x**4*exp_polar(I*pi)/a)/(sqrt(a)*gamma(11/4)) + e**4*x**9*gamma(9/4)*hyper((1/2, 9/4), (13
/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(13/4))

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